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Calculate ionisation constant for pyridinium hydrogen chloride. (Given that $\mathrm{H}^{+}$ion concentration is $3.6 \times 10^{-4} \mathrm{M}$ and its concentration is $0.02 \mathrm{M}$.)
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The correct answer is:
$1.5 \times 10^{-9}$
Pyridinium hydrochloride is a salt of weak base and strong acid.
$\begin{aligned} & \mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]=-\log \left(3.6 \times 10^{-4}\right)=3.44 \\ & \text { Now, } \mathrm{pH}=-\frac{1}{2}\left[\log K_w-\log K_a+\log C\right) \\ & \Rightarrow \quad 3.44=-\frac{1}{2}\left[-14-\log K_a+\log \left(2 \times 10^{-2}\right)\right] \\ & \Rightarrow 6.88=14+\log K_a+1.70 \\ & \Rightarrow \log K_a=-8.82 \\ & \Rightarrow K_a=\operatorname{antilog}(-8.82)=1.5 \times 10^{-9}\end{aligned}$
$\begin{aligned} & \mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]=-\log \left(3.6 \times 10^{-4}\right)=3.44 \\ & \text { Now, } \mathrm{pH}=-\frac{1}{2}\left[\log K_w-\log K_a+\log C\right) \\ & \Rightarrow \quad 3.44=-\frac{1}{2}\left[-14-\log K_a+\log \left(2 \times 10^{-2}\right)\right] \\ & \Rightarrow 6.88=14+\log K_a+1.70 \\ & \Rightarrow \log K_a=-8.82 \\ & \Rightarrow K_a=\operatorname{antilog}(-8.82)=1.5 \times 10^{-9}\end{aligned}$
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